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KEAM 2015 Math Paper

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Question : 117 of 120

Marks: +1, -0

If

|t| < 1,

\sin x = \frac{2t}{1+t^2},

\tan y = \frac{2t}{1-t^2},

\frac{dy}{dx} =

$\frac{1}{x}$
$\frac{1}{2}$
$-\frac{1}{2}$
$-\frac{1}{x}$
1

Solution:

We have,

\sin x = \;\frac{2t}{1+t^{2}}

\Rightarrow \;\; x = \sin^{-1} \;\frac{2t}{1+t^{2}}

On putting

t = \tan \theta,

we get

x = \sin^{-1}\left( \frac{2 \tan \theta}{1+\tan^2 \theta} \right)

= \sin^{-1} \sin 2\theta = 2\theta

\Rightarrow \;\; x = 2 \tan^{-1} t

....(i)

Also,

\tan y = \;\frac{2t}{1-t^{2}}

\Rightarrow \;\; y = \tan^{-1} \;\frac{2t}{1-t^{2}}

On putting

t = \tan \theta,

we get

y = \tan^{-1}\left( \frac{2 \tan \theta}{1-\tan^2 \theta} \right)

= \tan^{-1}(\tan 2\theta) = 2\theta

= 2 \tan^{-1} t

....(ii)

From Eqs. (i) and (ii),

y = x

\Rightarrow \;\; \frac{dy}{dx} = 1

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